ch04-p048

# ch04-p048 - 48. Using the fact that v y = 0 when the player...

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Substituting the above expression into Eq. 4-22, we find the maximum height to be 2 22 2 00 0 0 max 0 0 max max 0 0 sin sin sin 11 (s i n) s i n . 2 vv v yv t g t v g gg g θθ θ §· =− = = ¨¸ ©¹ To find the time when the player is at max /2 yy = , we solve the quadratic equation given in Eq. 4-22: 2 0 0 max 0 0 sin (2 2) sin i . 24 2 2 v tg t t ± ± == = ¡ = With tt = (for ascending), the amount of time the player spends at a height max is m a x max max sin (2 sin sin 1 0.707 2 2 v t t t t g Δ Δ= − = = = ¡ . Therefore, the player spends about 70.7% of the time in the upper half of the jump. Note that the ratio
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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